MHB Dominika's question at Yahoo Answers Regarding Convex Functions

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A function \( f \) is convex on the interval \( (a,b) \) if and only if its graph lies above or on the tangent line at any point \( (c, f(c)) \) within that interval. The proof begins by assuming \( f \) is convex and demonstrating that it satisfies the inequality \( f(x) \geq f(c) + f'(c)(x-c) \) for all \( x, c \in (a,b) \). Conversely, if \( f \) lies above all tangent lines, the proof shows that this leads to the definition of convexity. The discussion emphasizes the equivalence of the geometric interpretation of convexity with the mathematical condition involving derivatives. Understanding this relationship is crucial for analyzing the properties of differentiable functions.
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Here is the question:

Dominika said:
Let $f$ be differentiable on the interval $(a,b)$. Show that $f$ is convex on $(a,b)$ if and only if, for every $c$ from $(a,b)$ its graph is placed above or on the tangent passing through a point $(c,f(c))$
Thanks a lot..

Here is a link to the question:

Convex function ...? - Yahoo Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hi Dominika,

The statement we want to prove is the following:

Suppose that $f$ is differentiable on it's domain $(a,b)$. Then $f$ is convex if and only if $(a,b)$ is convex and $f(x)\geq f(c)+f^{\prime}(c)(x-c)$ for all $x,c\in (a,b)$.

We're going to assume (without proof) the fact that an interval is convex (i.e. for any $x,y\in (a,b)$, we have $ty+(1-t)x\in (a,b)$ for any $0\leq t\leq 1$).

We proceed with the proof as follows.

Proof: Suppose $f$ is convex on $(a,b)$. Then for any $t\in[0,1]$ and $x,c\in(a,b)$, $tx + (1-t)c = c + t(x-c) \in (a,b)$. Furthermore, by definition of convexity of $f$, we have that
\[f(tx+(1-t)c) = f(c+t(x-c)) \leq tf(x) + (1-t)f(c).\]
Dividing both sides by $t$ and then solving the inequality for $f(x)$ gives us
\[f(x) \geq f(c) + \frac{f(c+t(x-c))-f(c)}{t}\]
Taking the limit as $t\to 0$ gives us
\[f(x)\geq f(c) + f^{\prime}(c)(x-c),\tag{1}\]
which is the desired result ($f(x)\geq\text{tangent line}$).

Conversely, suppose that $f$ lies above any tangent line over the interval $(a,b)$. Choose $x,c\in(a,b)$ such that $x\neq c$. Now, for any $t\in [0,1]$, define $z=tx + (1-t)c= c+t(x-c)$. We can see from here that $\color{red}{x-z}= \color{red}{(1-t)(x-c)}$ and $\color{blue}{c-z}=\color{blue}{-t(x-c)}$. Applying $(1)$ twice gives us
\[f(x)\geq f(z) + f^{\prime}(z)(x-z)\tag{2}\]
and
\[f(c)\geq f(z)+f^{\prime}(z)(c-z).\tag{3}\]
If we now multiply $(2)$ by $t$ and $(3)$ by $(1-t)$ and add them together, we see that
\[\begin{aligned} tf(x)+(1-t)f(c) &\geq tf(z)+(1-t)f(z) + tf^{\prime}(z)(\color{red}{x-z}) + (1-t)f^{\prime}(z)(\color{blue}{c-z})\\ &= f(z) + tf^{\prime}(z)\color{red}{(1-t)(x-c)} + (1-t)f^{\prime}(z)(\color{blue}{-t(x-c)})\\ &= f(z),\end{aligned}\]
which is the condition of convexity. Therefore, $f$ is convex.$\hspace{.25in}\blacksquare$

I hope this makes sense! (Smile)
 
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